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\title{Additional Notes of Regime 3 ($B>0$  $j>0$)}
\author{Xinya Zhang}
\date{\today}

\begin{document}

\maketitle

\section{Purpose of this Notes}
When I used the equations (4), (7), (24), and (25) to solve this
regime, I found the numerical results is quite unacceptable: $k$ and
$h$ are decreasing, and $j$ is negative. Plus, the equation (22) is
not satisfied. I think it may be because that differentiation from
(22) to (23) lose some information and enlarge the solution set
improperly, and the computer converges to the wrong solution. So I
tried to find some other way to solve the problem.

(Note: All the equation numbers in the above paragraph are based on Professor Hartley's notes)

\section{Hamiltonian System}
For the easy-read purpose, I copy all the first order equations and
differential equations of co-state variables here, and I will use the
equation numbers listed below from now on:
\begin{eqnarray}
\frac{\partial \mathcal{H}}{\partial c} &=& c^{-\gamma} -\lambda = 0
\label{eq:FOCc}  \\
\frac{\partial \mathcal{H}}{\partial i} &=& -\lambda + q = 0 \label{eq:FOCi}\\
\frac{\partial \mathcal{H}}{\partial j} &=& -\lambda + \eta + \mu = 0; \mu j=0, \mu\ge 0, j \ge 0 \label{eq:FOCj}\\
\frac{\partial \mathcal{H}}{\partial n} &=& -\lambda +\nu+\omega = 0, \omega n=0, \omega\ge 0, n\ge 0  \label{eq:FOCn}\\
\frac{\partial \mathcal{H}}{\partial R} &=& -\lambda g(S,N)+\epsilon+\sigma Q+\xi = 0, \xi R=0, \xi\ge 0, R\ge 0 \label{eq:FOCR} \\
\frac{\partial \mathcal{H}}{\partial B} &=& -\lambda \bigl (\Gamma_0+(\Gamma_1+H)^{-\alpha}\bigr )+\epsilon+\eta\psi+\zeta=0, \zeta B=0, \zeta\ge 0, B\ge 0 \label{eq:FOCB}
\end{eqnarray}
The differential equations:
\begin{eqnarray}
\dot{q} &=& \beta q-\frac{\partial \mathcal{H}}{\partial k} = (\beta+\delta) q - \lambda A+\epsilon A \label{eq:qdot}\\
\dot{\eta} &=& \beta \eta- \frac{\partial \mathcal{H}}{\partial H} =\beta\eta - \lambda\alpha (\Gamma_1+H)^{-\alpha-1}B \label{eq:etadot} \\
\dot{\sigma} &=& \beta \sigma - \frac{\partial \mathcal{H}}{\partial S} = \beta \sigma +\lambda \frac{\partial g}{\partial S}R  \label{eq:sigmadot} \\
\dot{\nu} &=& \beta \nu - \frac{\partial \mathcal{H}}{\partial N} = \beta\nu +\lambda \frac{\partial g}{\partial N}R  \label{eq:nudot}
\end{eqnarray}

I also put the budget constraint and the other useful equations here:
\begin{eqnarray}
 y &=& c+i+j+n+g(S,N)R+pB \label{eq:Budget} \\
 p &=& \Gamma_0+(\Gamma_1+H)^{-\alpha} \label{eq:RenewCost} \\
\dot{k} &=& i-\delta{k} \label{eq:kdot} \\
\dot{H} &=& j+\psi B \label{eq:Hdot}
\end{eqnarray}

\section{Differential Equation System of Regime 3}

In Regime 3, since we begin to use backstop technology, we have $R=0$,
$n=0$, $B>0$, $j>0$. At $T_2$, energy supply switches to the backstop
technology, while after $T_3$, $j$ becomes zero.

Since $B>0$, we have its multiplier $\zeta=0$. from FOC condition (6)
, we obtain
\begin{equation}
\epsilon=\lambda(\Gamma_0+(\Gamma_1+H)^{-\alpha})-\eta\psi
\end{equation}

Also we have $j>0$, and then $\mu=0$. From FOC conditon (2) and (3),
we know $\lambda=q=\eta$, and $\dot{q}=\dot{\eta}$.  Combine with
differential equations (7) and (8), and subtitute (15) into (7), we
will get
\begin{equation}
(\Gamma_1+H)^{-\alpha}+\alpha k(\Gamma_1+H)^{-\alpha-1} = 1+ \psi-\Gamma_0-\frac{\delta}{A}\label{eq:qdot_eq_etadot}
\end{equation}
Thus, we can write $k$ as an expression of $H$:
\begin{eqnarray}
k=\frac{\bar{B}-(\Gamma_1+H)^{-\alpha}}{\alpha(\Gamma_1+H)^{-\alpha-1}} \\
where\; \bar{B} = 1+\psi-\Gamma_0-\delta/A \nonumber
\end{eqnarray}

Differentiating (16) with respect to t, we find:
\begin{equation}
\dot{k}-\dot{H}-(1+\alpha)k\dot{H}/(\Gamma_1+H)=0
\label{eq:kdot_Hdot}
\end{equation}

Similarly, we can rewrite (18) as:
\begin{equation}
\dot{k} = \dot{H}+\frac{(1+\alpha)k\dot{H}}{\Gamma_1+H}
\\ = \dot{H}+\frac{(1+\alpha)\dot{H}}{\Gamma_1+H}\cdot\frac{\bar{B}-(\Gamma_1+H)^
{-\alpha}} {\alpha(\Gamma_1+H)^{-\alpha-1}}
\end{equation}
which means that $\dot{k}$ can be written as a formula of $H$ and $\dot{H}$ .

From (13) and (14), we know that $i=\dot{k}+\delta k$ and
$j=\dot{H}-\psi B$. Substitute these two equations into budget
constraint (11), we have
\begin{equation}
c+\dot{k}+\delta k+\dot{H}-\psi B+pB=y
\end{equation}
Note that in this regime, $B=y=Ak$, and by FOC condition (1),
$c=\lambda^{-1/\gamma}$. Then we have $\lambda$, $k$, $\dot{k}$, $H$,
and $\dot{H}$ in the equation (20). However, by using (17) and (19),
we can eliminate $k$ and $\dot{k}$, and only have $\lambda$, $H$, and
$\dot{H}$ in this equation:
\begin{equation}
\dot{H}\cdot(2+\frac{1+\alpha}{\Gamma_1+H}\cdot k)
=[A(1+\psi-p)-\delta]k-\lambda^{-1/\gamma}
\end{equation}
where according to (17),  $k=\frac{\bar{B}-(\Gamma_1+H)^{-\alpha}}{\alpha(\Gamma_1+H)^{-\alpha-1}}$.

From $\dot{\lambda}=\dot{q}=\dot{\eta}$, and equation (8), we obtain:
\begin{equation}
\dot{\lambda} = (\beta - \alpha Ak(\Gamma_1+H)^{-\alpha-1})\lambda \label{eq:q_eta_lambda}
\end{equation}


Given $H_{T_3}$, and $\lambda_{T_3}$, we can solve equation (21) and (22) backward.

I tried to solve them in the computer, and the results seems
reasonable, at least k and h are increasing. However, there might be
mistakes in the algebra.

\begin{eqnarray}
  \label{eq:sss}
  a=b+c
\end{eqnarray}



\end{document}
